A124093 -id:A124093 - cp's OEIS frontend

A123596 Squares alternating with triangular numbers.

0, 0, 1, 1, 4, 3, 9, 6, 16, 10, 25, 15, 36, 21, 49, 28, 64, 36, 81, 45, 100, 55, 121, 66, 144, 78, 169, 91, 196, 105, 225, 120, 256, 136, 289, 153, 324, 171, 361, 190, 400, 210, 441, 231, 484, 253, 529, 276, 576, 300, 625, 325, 676, 351, 729, 378, 784, 406, 841, 435

Offset: 0

Peter Hansen (babyskbaby(AT)web.de), Nov 14 2006

nonn

Rearrangement of A054686.

For n >= 2, a(n) is the number of distinct positive slopes of least squares regression lines fitted to n points (j,y_j), 1 <= j <= n, where all y_j are 0 or 1. The total number of distinct slopes is 2*a(n)+1 (a(n) positive, a(n) negative, and the zero slope). - Pontus von Brömssen, Mar 10 2024

Alois P. Heinz, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (0,3,0,-3,0,1).

Cf. A002620, A054686, A124093.

[(3*n^2-1+(n^2+1)*(-1)^n)/16: n in [0..10]]; // G. C. Greubel, Oct 26 2017

CoefficientList[Series[x^2*(1+x+x^2)/((1-x)^3*(1+x)^3), {x, 0, 50}], x] (* or *) Table[(3*n^2-1+(n^2+1)*(-1)^n)/16, {n,0,50}] (* G. C. Greubel, Oct 26 2017 *)
With[{nn=30},Riffle[Range[0,nn]^2,Accumulate[Range[0,nn]]]] (* or *) LinearRecurrence[{0,3,0,-3,0,1},{0,0,1,1,4,3},60] (* Harvey P. Dale, Feb 11 2020 *)

{a(n) = if(n%2,(n^2-1)/8,n^2/4)} \\ Michael Somos, Nov 18 2006

a(2*n) = n^2, a(2*n+1) = (n^2+n)/2.
From R. J. Mathar, Feb 12 2010: (Start)
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x^2*(1+x+x^2)/((1-x)^3*(1+x)^3). (End)
a(n) = (3*n^2-1+(n^2+1)*(-1)^n)/16. - Luce ETIENNE, May 30 2015

Edited by Michael Somos, and several other correspondents, Nov 14 2005

A267322 Expansion of (1 + x + x^2 + x^4 + 2*x^5)/(1 - x^3)^3.

Original entry on oeis.org

1, 1, 1, 3, 4, 5, 6, 9, 12, 10, 16, 22, 15, 25, 35, 21, 36, 51, 28, 49, 70, 36, 64, 92, 45, 81, 117, 55, 100, 145, 66, 121, 176, 78, 144, 210, 91, 169, 247, 105, 196, 287, 120, 225, 330, 136, 256, 376, 153, 289, 425, 171, 324, 477, 190, 361, 532, 210, 400, 590, 231, 441, 651

Offset: 0

Ilya Gutkovskiy, Apr 07 2016

Triangular numbers alternating with squares and pentagonal numbers.

			Illustration of initial terms:
==========================================================
n:    0   1   2     3     4     5       6       7       8
----------------------------------------------------------
                                                        o
                                                      o o
                                o       o   o o o   o o o
                    o   o o   o o     o o   o o o   o o o
      o   o   o   o o   o o   o o   o o o   o o o   o o o
==========================================================
      1   1   1     3     4     5       6       9      12
----------------------------------------------------------

Ilya Gutkovskiy, Extended illustration of initial terms
Eric Weisstein's World of Mathematics, Triangular Number
Eric Weisstein's World of Mathematics, Square Number
Eric Weisstein's World of Mathematics, Pentagonal Number
Index entries for linear recurrences with constant coefficients, signature (0,0,3,0,0,-3,0,0,1)

Cf. A000217, A000290, A000326, A123596, A124093, A271391.

LinearRecurrence[{0, 0, 3, 0, 0, -3, 0, 0, 1}, {1, 1, 1, 3, 4, 5, 6, 9, 12}, 70]
Table[(Floor[n/3] + 1) ((n + 1) Floor[n/3] - 3 Floor[n/3]^2 + 2)/2, {n, 0, 70}] (* Bruno Berselli, Apr 08 2016 *)
CoefficientList[Series[(1+x+x^2+x^4+2x^5)/(1-x^3)^3,{x,0,70}],x] (* Harvey P. Dale, Dec 31 2023 *)

x='x+O('x^99); Vec((1+x+x^2+x^4+2*x^5)/(1-x^3)^3) \\ Altug Alkan, Apr 07 2016

G.f.: (1 + x + x^2 + x^4 + 2*x^5)/(1 - x^3)^3.
a(n) = 3*a(n-3) - 3*a(n-6) + a(n-9).
a(3k) = A000217(k+1), a(3k+1) = A000290(k+1), a(3k+2) = A000326(k+1).
Sum_{n>=0} 1/a(n) = 2 - Pi/sqrt(3) + Pi^2/6 + 3*log(3) = 5.1269715686...
a(n) = (floor(n/3) + 1)*((n+1)*floor(n/3) - 3*floor(n/3)^2 + 2)/2. - Bruno Berselli, Apr 08 2016

cp's OEIS Frontend

A123596 Squares alternating with triangular numbers.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions